IMO 1J, Independent Study 1

By the weighted AM-GM inequality we have
$$\begin{aligned} \frac{2x \times y + 2 \times (2-y)}{y + (2-y)} &\ge \sqrt{(2x)^y \times 2^{2-y}} \\ \Rightarrow xy + (2-y) &\ge \sqrt{4x^y} \\ \Rightarrow (xy + 2 - y)^2 &\ge 4x^y. \end{aligned}$$
Similarly we have
$$(xy + 2 - x)^2 \ge 4y^x.$$
Adding the two expressions gives
$$\begin{aligned} 4x^y + 4y^x &\le (xy + 2 - y)^2 + (xy + 2 - x)^2 \\ &= 2(xy + 2)^2 - 2(x + y)(xy + 2) + x^2 + y^2 \\ &= 2 \{ (xy)^2 + 4xy + 4 \} - \{ 4xy + 8 \} + (x + y)^2 - 2xy \\ &= 2(xy)^2 + 2xy + 4. \end{aligned}$$

Dividing both sides of the inequality by 2 completes the proof.